Let $F:[0,1] \rightarrow [0,1]$ belong to $C^2([0,1])$ and verifies $F'(x)>0$ and $F''(x)>0$ for all $x \in (0,1)$.
Show that $F'(0) < 1$ (that the origin is an attractor point)
What I've come so far:
Expand $F'(x)$ through Taylor series around $0$:
$$F'(x) = F'(0)+F''(0)(x-0)$$ $$ F''(0) = \frac{F'(x)-F'(0)}{x}$$ Then if $F''(x)>0$ $\forall x \in (0,1)$: $$\frac{F'(x)-F'(0)}{x} > 0$$ $$F'(x) > F'(0)\space \space\forall x \in (0,1)$$ Assume $F'(0) \geq 1$ then $F'(x) > 1 \space \space \forall x \in (0,1)$
Expand $F(x)$ through Taylor series around $0$: $$ F(x) = F(0)+F'(0)(x-0)$$ $$F'(0) = \frac{F(x)-F(0)}{x} > 1$$ $$F(x) > x + F(0)$$ But since $F(x)$ must be bounded below by $x+F(0)$ and $F(x) \in [0,1]$, $x+F(0) \neq 0$ it would imply that $F(x) \in (0,1]$, so $F'(0)>1$
I'm not sure about this last argument which I think it's incorrect since assuming that $F'(0)<1$ I would get an upper bound of $F(x)$ with which I could use the same argument to refuse that $F'(0)<1$...
As mentioned in the comments, $$ F'(x) = F'(0)+F''(0)(x-0) $$ is wrong (unless $F'$ is linear). What you have (from the mean-value theorem) is that $$ F'(x) = F'(0)+F''(c)(x-0) $$ for some $c \in (0,x)$. For the same reason, $$ F(x) = F(0)+F'(0)(x-0) $$ is not correct in general.
But you can proceed as follows: Taylor's formula for $F$ on the interval $[0, 1]$ gives $$ 1 \ge F(1) = F(0) + (1-0) F'(0) + \frac 12 (1-0)^2 F''(c) \ge F'(0) + \frac 12 F''(c) > F'(0) $$ for some $c \in (0,1)$.
Alternatively, use the mean-value theorem to show that $$ 1 \ge \frac{F(1)-F(0)}{1-0} = F'(c) $$ for some $c \in (0,1)$, and that $F'$ is strictly increasing.